Here is a sangaku - a Temple geometry - problem in an equilateral triangle similar to another one in a square. The similarity aside, the current one is much more difficult than the other.
This is the problem #2.1.7 from [Fukagawa and Pedoe]. The problem is from an existent 1886 tablet found in the Fukusima prefecture: In an equilateral triangle ABC the lines AC'A', BA'B', CB'C' are drawn making equal angles with AB, BC, CA, respectively, forming the triangle A'B'C', and so that the radius of the incircle of triangle A'B'C' is equal to the radius of the incircle of triangle AA'B. Find A'B' in terms of AB.
Unhappily I must confess that the equations are am able to write appear unappealing. I just could not see any shortcuts that would lead to an elegant solution.
Having no solution, I do not know whether the following observation is relevant or not: lines AA', BB', CC' touch, by the construction three of the four incircles. When the four have equal radii, the lines parallel to the former but passing on the other side of the incircle of ΔA'B'C' seem to also touch three of the four circles.